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## Definition

Factorization is the decomposition of an expression into a product of its factors.

The following are common factorizations.

1. For any positive integer $$n$$, $a^n-b^n = (a-b)(a^{n-1} + a^{n-2} b + \ldots + ab^{n-2} + b^{n-1} ).$ In particular, for $$n=2$$, we have $$a^2-b^2=(a-b)(a+b)$$.

2. For $$n$$ an odd positive integer, $a^n+b^n = (a+b)(a^{n-1} - a^{n-2} b + \ldots - ab^{n-2} + b^{n-1} ).$

3. $$a^2 \pm 2ab + b^2 = (a\pm b)^2$$

4. $$x^3 + y^3 + z^3 - 3 xyz = (x+y+z) (x^2+y^2+z^2-xy-yz-zx)$$

5. $$(ax+by)^2 + (ay-bx)^2 = (a^2+b^2)(x^2+y^2)$$. $$(ax-by)^2 - (ay-bx)^2 = (a^2-b^2)(x^2-y^2)$$.

6. $$x^2 y + y^2 z + z^2 x + x^2 z + y^2 x + z^2 y +2xyz= (x+y)(y+z)(z+x)$$.

Factorization often transforms an expression into a form that is more easily manipulated algebraically, that has easily recognizable solutions, and that gives rise to clearly defined relationships.

## Worked Examples

### 1. Find all ordered pairs of positive integer solutions $$(x,y)$$ such that $$2^x+ 1 = y^2$$.

Solution: We have $$2^x = y^2-1 = (y-1)(y+1)$$. Since the factors $$(y-1)$$ and $$(y+1)$$ on the right hand side are integers whose product is a power of 2, both $$(y-1)$$ and $$(y+1)$$ must be powers of 2. Furthermore, their difference is

$(y+1)-(y-1)=2,$

implying the factors must be $$y+1 = 4$$ and $$y-1 = 2$$. This gives $$y=3$$, and thus $$x=3$$. Therefore, $$(3, 3)$$ is the only solution.

### 2. Factorize the polynomial

$f(a, b, c) = ab(a^2-b^2) + bc(b^2-c^2) + ca(c^2-a^2).$

Solution: Observe that if $$a=b$$, then $$f(a, a, c) =0$$; if $$b=c$$, then $$f(a, b, b)=0$$; and if $$c=a$$, then $$f(c,b,c)=0$$. By the Remainder-Factor Theorem, $$(a-b), (b-c),$$ and $$(c-a)$$ are factors of $$f(a,b,c)$$. This allows us to factorize

$f(a,b,c) = -(a-b)(b-c)(c-a)(a+b+c).$

Note by Calvin Lin
2 years, 7 months ago

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For the worked example - 1 : there are 2 solutions : (3,3) and (3,-3)

(y+1)-(y-1) = 2

implies y + 1 = -2 or y + 1 = 4 and y -1 = -4 or y-1 = 2

Thus, the two solutions you have are (3,3) and (3,-3) · 2 years, 4 months ago

Thanks. I added in "positive integers". Staff · 1 year, 4 months ago

I did not understand why should (y+1) - (y-1) = 2 ?
Can anyone explain ? · 1 year, 4 months ago

What do you not understand?

What do you think $$(y+1) - (y-1)$$ is equal to ? Staff · 1 year, 4 months ago

I did not see it correctly . My fault ! Sir, I have a problem. I want to learn Number theory as is organized here on Brilliant but I don't follow anything beginning from modular inverses.I have tried the wikis but I still don't follow .Please suggest something. · 1 year, 4 months ago

Thanks for this. · 2 years, 4 months ago

How to think that both difference is 2. What is the main moto behind thinking such that???? · 2 years, 4 months ago

How is their (y+1)-(y-1) difference 2? · 2 years, 4 months ago

$(y+1) - (y-1)$ $= y +1 - y + 1$ $= 1+1= \boxed{2}$ · 2 years, 4 months ago

(Y+1)-(y-1) = y+1-y+1= 2 · 2 years, 4 months ago